Let K be a field and G a finite group. We know that when char(K) does not divide |G|, the group algebra K[G] is semisimple. Conversely we have:

Proposition. If char(K) divides |G|, then K[G] is not semisimple.

Proof

Let , a two-sided ideal of K[G]. If K[G] = I⊕J for some left ideal J of K[G], then dim J = 1 so it is spanned by, say, . Since αg is a scalar multiple of α for each g∈G, all the bg‘s must be equal to some b∈K. But this means lies in I since its sum of coefficients is ♦

For the case where char(K) divides |G|, we have modular representationt theory. Here, we’ll require some knowledge of complete discrete valuation rings; throughout this article, we adopt the following notations and assumptions.

R is a complete discrete valuation ring with maximal ideal generated by π∈R.

K is the field of fractions of R.

k := R/(π) is the residue field.

p := char(k) divides |G| and char(K) = 0.

Representations of G over k are studied by first lifting them to R, then projecting to K. Let be the Grothendieck group of finitely-generated modules over k[G], K[G] respectively. Since char(K) = 0, K[G] is semisimple. Since k[G] is of finite dimension over k, it is artinian; we let J be its Jacobian radical. Finally, let be the Grothendieck group of finitely-generated projective modules over k[G]. From earlierresults:

are free abelian groups, with the following bases, which we will fix from now on.

The pairing taking is a bi-additive map. The above basis for is orthogonal by Schur’s lemma. Extending K if necessary, we may assume it is orthonormal.

The pairing taking is a bi-additive map. The given bases for and are dual with respect to this pairing. [ Indeed, for with M simple, we have Hom(P, M) = Hom(P/JP, M) since JM = 0; if P is projective indecomposable, then P/JP is simple and the result follows from Schur’s lemma. Assuming k is large enough, we may assume that if P/JP ≅ M. ]

The next step is to define the maps d and e in the following diagram, such that c = de:

Definition of d

To define , let M be a finitely-generated K[G]-module.

Proposition. There is an R[G]-module N such that . The class of this module depends only on M.

Proof

Step 1. First prove the existence of N.

Let N’ ⊂ M be a free R-module with a basis spanning M, so as K-vector spaces. Now let which is also a free R-module since it is torsion-free, and has a basis spanning M. Thus Since N is G-invariant it is an R[G]-module.

Step 2. Proof of uniqueness: initial step.

Suppose are R[G]-modules such that Replacing by a scalar multiple (this does not affect ), we may assume Now for each we have for some r; since has a finite basis, we have for some r>0.

Step 3. Now we show: if , then in

This is by induction on r. When r=1, let ; we have an exact sequence of k[G]-modules:

Since we have in the group For r>1, let we have:

By induction hypothesis, and we’re done. ♦

We thus define via: given pick an R[G]-module N such that We then define

Note

An exact sequence of K[G]-modules splits as M ≅ M’ ⊕ M”; picking modules N’ and N” for M’ and M” via the above proposition, N := N’ ⊕ N” also satisfies the proposition for M, and we obtain d([M]) = d([M’]) + d([M”]) as desired, so d is well-defined.

Indeed, they’re isomorphic over K[G] since the above two matrices are diagonalizable with eigenvalues -1, +1. However, reducing modulo 2 gives non-isomorphic k[G]-modules:

Note that they have the same composition factors, although one is semisimple and the other is not.

Definition of e

Next, we will define . This is given in two steps: first we consider the map , where is the Grothendieck group of finitely-generated projective R[G]-modules. [ Warning: since R[G] is not artinian, we have to avoid using results from the previous two articles. ] Note that Ψ is well-defined:

If P is projective over R[G], then for some n so and so P/πP is projective over k[G].

An exact sequence of projective modules must split, so Ψ is a well-defined homomorphism of abelian groups.

The next step is to show that Ψ is an isomorphism. First, injectivity:

Lemma. If P, Q are finitely-generated and projective such that as k[G]-modules, then as R[G]-modules.

Proof

Let be an isomorphism of k[G]-modules. Since P is projective, the map lifts to a map of R[G]-modules such that the reduction of f1 mod π is f0. To show that f1 is bijective, it suffices to show it is an isomorphism of R-modules. But P, Q are finite R-free since they are projective over R[G] and thus torsion-free. So we only need to show that det(f1) is invertible in R, i.e. ≠0 modulo π. But det(f1) mod π = det(f0) ≠ 0. ♦

Hence, is injective. Finally, we have the following:

Lemma. For any finitely-generated projective k[G]-module N, there is a finitely-generated projective R[G]-module M such that

Proof

We may assume N is indecomposable, so N is a direct summand of k[G], i.e. The image of 1∈k[G] in N then gives an idempotent x∈N (i.e. x2=x). We wish to lift x to an idempotent y∈R[G] such that x = y mod π.

Claim: if S is any ring and I ⊂ S is an ideal satisfying I2=0, then any idempotent y ∈ S/I lifts to an idempotent x ∈ S.

Now consider the ring with ideal This satisfies I2 = 0 and S/I ≅ k[G] so by the claim, x lifts to an idempotent y’∈S.

Repeat the process with the ring with ideal Again and S’/I’ ≅ S so y’ lifts to an idempotent y”∈S’. Repeating this process, since R is a complete discrete valuation ring, we obtain an idempotent y∈R[G] such that x = y mod π.

Then R[G] is the direct sum of left modules M := R[G]y and M’ := R[G](1-y), and ♦

Thus, Ψ is an isomorphism and we can define:

where the second map takes to

Relations Between d and e

Finally we wish to prove that c = de. Indeed, let and so that Then the element e([P]) is given by To obtain we recover Q as the R[G]-module in the definition of d and take: as desired.

other : take the subspace of K3 given by (x, y, z) satisfying x+y+z = 0; G acts by permuting the coordinates.

This gives a basis of . On the other hand, let’s consider all simple k[G]-modules. We still have the trivial and alternating representations, clearly simple. But now the third representation has a submodule:

since char(k) = 3. This gives a subspace which is the trivial representation, and the quotient is alternating. Hence, has two simple modules, given by the trivial and alternating modules, and the matrix for is:

Next we compute e; to do that we need to find the indecomposable projective k[G]-modules. These can be found by decomposing k[G] itself. From above, we see that the key is to find idempotents of k[G] (specifically indecomposable projectives correspond to primitive idempotents, but we won’t use that here). Clearly is idempotent, and we obtain:

We leave it to the reader to prove that these are indecomposable. To compute e([W]) and e([W’]), we lift them to projective R[G]-modules then tensor with K: this still gives

which has character values (3, 1, 0) and (3, -1, 0), i.e. and So the matrix for e is: